determinant inequalities

1) |det⁡(A)|≤ρn⁢(A)2) |det⁡(A)|≤∏i=1n(∑j=1n|ai⁢j|)=∏i=1n∥ai∥13) |det⁡(A)|≤∏j=1n(∑i=1n|ai⁢j|)=∏j=1n∥aj∥14) |det⁡(A)|≤∏i=1n(∑j=1n|ai⁢j|2)12=∏i=1n∥ai∥25) |det⁡(A)|≤∏j=1n(∑i=1n|ai⁢j|2)12=∏j=1n∥aj∥26) if A is Hermitianpositive semidefinite, det⁡(A)≤∏i=1nai⁢i, with equality if and only if A is diagonal.

(Note that inequalities 2)-5) may suggest the idea that such inequalities could hold: |det⁡(A)|≤∏i=1n∥ai∥p or |det⁡(A)|≤∏j=1n∥aj∥p for any p∈𝐍; however, this is not true, as one can easily see with A=[11-11] and p=3. Actually, inequalities 2)-5) give the best possible estimate of this kind.)

Proofs:

1) |det⁡(A)|=|∏i=1nλi|=∏i=1n|λi|≤∏i=1nρ⁢(A)=ρn⁢(A).

2) If A is singular, the thesis is trivial. Let then det⁡(A)≠0. Let’s define B=D⁢A, D=d⁢i⁢a⁢g⁢(d11,d22,⋯,dn⁢n),di⁢i=(∑j=1n|ai⁢j|)-1. (Note that di⁢i exist for any i, because det⁡(A)≠0 implies no all-zero row exists.) So ∥B∥∞=maxi⁡(∑j=1n|bi⁢j|)=1 and, since ρ⁢(B)≤∥B∥∞, we have: